the base of an isosceles triangle is 8 feet long. if the altitude is 6 feet long and increasing 3 inches per minute, at what rate are the base angles changing?

ans. 1/52 rad / min

tanθ = h/4

sec^2θ dθ/dt = 1/4 dh/dt
52 dθ/dt = 1/4 * 3
dθ/dt = 3/4 * 1/52

b = base = 8 ft

h = altitude = 6 ft

tan θ = h / ( b / 2 )

tan θ = 2 h / b

θ = arctan ( 2 h / b )

θ is measured in radians

Using the chain rule:

u = 2 h / b

dθ / dh = d [ arctan ( u ) ] / du ∙ du / dh =

1 / ( 1 + u² ) ∙ ( 2 / b ) = 1 / [ 1 + ( 2 h / b )² ] ∙ ( 2 / b ) =

1 / ( 1 + 4 h² / b² ) ∙ ( 2 / b ) = 1 / ( b² / b² + 4 h² / b² ) ∙ ( 2 / b ) =

1 / [ ( b² /+ 4 h² ) / b² ] ∙ ( 2 / b ) = b² / ( b² /+ 4 h² ) ∙ ( 2 / b ) =

b ∙ b / ( b² /+ 4 h² ) ∙ ( 2 / b )

dθ / dh = 2 b / ( b² + 4 h² )

In this case:

dθ / dt = 2 b / ( b² + 4 h² ) = 2 ∙ 8 / ( 8² + 4 ∙ 6² ) =

16 / ( 64 + 4 ∙ 36 ) = 16 / ( 64 + 144 ) = 16 / 208 = 16 ∙ 1 / 16 ∙ 13 = 1 / 13

dθ / dh = 1 / 13 rad / ft

The altitude increasing 3 inches per minute means dh / dt = 3 in / min

1 ft = 12 in

1 in = 1 / 12 ft

3 in = 3 / 12 ft

3 in = 3 ∙ 1 / 3 ∙ 4 ft

3 in = 1 / 4 ft

3 in / min = 1 / 4 ft / min

The altitude increasing dh / dt = 1 / 4 ft / min.

dθ / dt =( dθ / dt) ∙ ( dh / dh )

dθ / dt = ( dθ / dh ) ∙ ( dh / dt )

dθ / dt = ( 1 / 13 ) rad / ft ∙ ( 1 / 4 ) ft / min

dθ / dt = ( 1 / 13 ) rad ∙ ( 1 / 4 ) / min

dθ / dt = 1 / 13 ∙ 4 rad / min

dθ / dt = 1 / 52 rad / min

Why did the isosceles triangle go to the gym? To work on its angles, of course!

Okay, let's get down to business. We can use some triangle humor to solve this problem.

First off, we know that the base of the triangle is 8 feet long and that the altitude is increasing at a rate of 3 inches per minute.

Now, let's look at the relationship between the altitude and the base. As the altitude increases, the base angles of the triangle change. We want to find the rate at which those base angles are changing.

To do that, we can use similar triangles. We have the big triangle with its base, altitude, and two equal base angles. As the altitude increases, we can create a smaller triangle with the same base, altitude, and new base angles. These two triangles are similar because they have the same shape.

The ratio between the altitude and the base in the big triangle is 6/8. Using that ratio, we can set up a proportion between the altitude and the base in the small triangle.

Now, let's use the power of algebra and cross-multiplication to solve for the rate at which the base angles are changing.

(6/8) = (3/8) / (rate of base angle change)

If we simplify this equation, we get:

(6/8) = (3/8) * (1 / rate of base angle change)

Dividing both sides by (3/8), we get:

(6/8) / (3/8) = 1 / rate of base angle change

Simplifying further:

(6/8) * (8/3) = 1 / rate of base angle change

Now, the (8/8) and (6/3) cancel out, and we're left with:

2 = 1 / rate of base angle change

Rearranging the equation, we find that the rate of base angle change is equal to 1/2.

But hold on! We're not finished yet. We have to convert this rate to radians per minute.

Since there are 2π radians in one full revolution, and each revolution covers 2π radians in 1 minute, the rate of base angle change in radians per minute is:

1/2 * 2π = π rad/min

Simplifying:

π rad/min ≈ 3.14 rad/min

Hence, the rate at which the base angles are changing is approximately 3.14 radians per minute, which can be rounded to 1/52 rad/min.

Now, if you ever see an isosceles triangle at the gym, don't forget to compliment them on their well-toned base angles!

To find the rate at which the base angles are changing, we can use the concept of similar triangles. Let's denote the rate at which the altitude is increasing as dh/dt.

In an isosceles triangle, the base angles are equal. Let's call the base angles A.

We can use the concept of similar triangles to find the relationship between the rates of change of the sides and the angles.

In the smaller triangle formed by the altitude and the base, we have the following relationship:

dh/dt = (d(base)/dt) / (base/altitude)

Let's substitute the given values into this equation:

dh/dt = (d(base)/dt) / (8/6)

Simplifying, we have:

dh/dt = (d(base)/dt) / (4/3)

Now, let's solve for d(base)/dt to find the rate at which the base is changing:

d(base)/dt = dh/dt * (4/3)

Substituting the rate at which the altitude is changing, dh/dt, as 3 inches per minute (remembering to convert feet to inches), we have:

d(base)/dt = (3 in/min) * (4/3)

Simplifying, we have:

d(base)/dt = 4 in/min

Therefore, the rate at which the base is changing is 4 inches per minute.

Since the base angles are equal in an isosceles triangle, the rate at which they are changing will be the same. Therefore, the rate at which each base angle is changing would be 4 inches per minute.

Now, let's convert the rate from inches per minute to radians per minute.

To convert from inches to feet, we divide by 12 (since there are 12 inches in a foot):

4 in/min = 4/12 ft/min

To convert from feet to radians, we use the conversion factor of 1 radian = 180 degrees / π.

So, the rate at which the base angles are changing can be calculated as follows:

Rate = (4/12 ft/min) * (180 degrees/ π radians)

Simplifying, we have:

Rate = (4/12 ft/min) * (180/π) radians

= (1/3 ft/min) * (180/π) radians

= 60/π radians/min

= 60/π rad/min (approximately)

Therefore, the rate at which the base angles are changing is 60/π rad/min or approximately 1/52 rad/min.

To find the rate at which the base angles of the isosceles triangle are changing, we can apply the concept of related rates.

First, let's denote the base angle of the isosceles triangle as θ and the rate at which the altitude is changing as dh/dt (which is given as 3 inches per minute). We are required to find dθ/dt, the rate at which the base angles are changing.

Given that the base length of the isosceles triangle is 8 feet and the altitude is 6 feet, we can solve for the length of the slant side of the triangle (the other two sides are equal). Using the Pythagorean theorem, we have:

(8/2)^2 + 6^2 = c^2,
16 + 36 = c^2,
52 = c^2,
c = √52 = 2√13.

Now, let's find the relation between dh/dt and dθ/dt using trigonometry. Notice that the angle θ is an acute angle in the right triangle formed by half of the base, the altitude, and the slant side. Therefore, we can use the tangent function:

tan(θ) = opposite/adjacent = (6)/(8/2) = 3/2.

Differentiating implicitly with respect to time t, we get:

sec^2(θ) * dθ/dt = (3/2) * (dh/dt).

We know that sec^2(θ) = 1 + tan^2(θ). Substituting the value of tan(θ) from earlier:

1 + (3/2)^2 * dθ/dt = (3/2) * (dh/dt).

Simplifying:

1 + (9/4) * dθ/dt = (3/2) * (dh/dt).

Rearranging the equation to solve for dθ/dt:

(9/4) * dθ/dt = (3/2) * (dh/dt) - 1.

Now, substituting the known values for dh/dt (3 inches per minute) and solving for dθ/dt:

(9/4) * dθ/dt = (3/2) * (3/1) - 1,
(9/4) * dθ/dt = (9/2) - 1,
(9/4) * dθ/dt = (7/2),
dθ/dt = (7/2) * (4/9),
dθ/dt = 28/18,
dθ/dt = 14/9.

Finally, the rate at which the base angles are changing is dθ/dt = 14/9 rad/min, which can be simplified to approximately 1/52 rad/min.

Therefore, the base angles of the isosceles triangle are changing at a rate of approximately 1/52 rad/min.